How to compute $sum_{n=1}^infty{frac{n}{(2n+1)!}}$?

Question

In a calculus book I am reading I have encountered the following problem:
$$sum_{n=1}^infty{frac{n}{(2n+1)!}}$$
The hint is to use Taylor series expansion's for $e^x$. I tried to express the sum as the form $$e^x=sum_{n=0}^{infty}{frac{x^n}{n!}}$$
But I could not find a consistent method, I always end un with different sums of factorials that does not help me solve the problem
The official solution is $$boxed{frac{1}{2e}}$$
The excersise is in a chapter that mixes calculus with summation, so the solution will probably include both.
Any help or hint is highly appreaciated! Thanks in advance.

Mark Viola · Accepted Answer

METHODOLOGY $1$:
The Taylor series for the hyperbolic sine function $sinh(x)$ is given by
$$sinh(x)=sum_{n=0}^infty frac{x^{2n+1}}{(2n+1)!}tag1$$
If we divide both sides of $(1)$ by $x$, differentiate, and set $x=1$, awe find that
$$underbrace{cosh(1)-sinh(1)}_{=e^{-1}}=2sum_{n=1}^inftyfrac{n}{(2n+1)!}tag2$$
Finally, dividing $(2)$ by $2$ yields the coveted result
$$sum_{n=1}^infty frac{n}{(2n+1)!}=frac1{2e}$$
as was to be shown!

METHODOLOGY $2$:
We begin with the Taylor series for $e^x$ at $x=-1$.  Then, we see that
$$begin{align}
frac1e&=sum_{n=0}^infty frac{(-1)^n}{n!}\
&=sum_{n=0}^inftyleft(frac1{(2n)!}-frac1{(2n+1)!}right)\
&=sum_{n=0}^infty frac{(2n+1)!-(2n)!}{(2n)!(2n+1)!}\
&=sum_{n=1}^infty frac{2n}{(2n+1)!}\
&=2sum_{n=1}^infty frac{n}{(2n+1)!}
end{align}$$
from which we arrive at the coveted result
$$sum_{n=1}^infty frac{n}{(2n+1)!}=frac1{2e}$$
as expected!

vonbrand · Answer

First off, this looks like the derivative of the following sum, evaluated at $x = 1$:
$begin{align*}
   &sum_{n ge 1} frac{x^n}{(2 n + 1)!}
end{align*}$
Trouble is that we get only odd terms of something like:
$begin{align*}
   e^{x^{1/2}}
     &= sum_{n ge 0} frac{x^{n/2}}{n!}
end{align*}$
Now if:
$begin{align*}
   f(z)
     &= sum_{n ge 0} a_n z^n
end{align*}$
then:
$begin{align*}
   frac{f(z) + f(-z)}{2}
      &= sum_{n ge 0} a_{2 n} z^{2 n} \
   frac{f(z) - f(-z)}{2}
      &= sum_{n ge 0} a_{2 n + 1} z^{2 n + 1}
end{align*}$
So your sum is:
$begin{align*}
  S(x)
     &= x^{1/2} frac{e^{x/2} - e^{-x/2}}{2} \
     &= sum_{n ge 0} frac{x^n}{(2 n + 1)!}
end{align*}$
What is left is routine:
$begin{align*}
   S'(1)
     &= frac{e}{2} 
end{align*}$
The manipulations are valid inside the region of convergence of the series for $e^x$, i.e., all of $mathbb{R}$.

Alex · Answer

Hint: $frac{n}{(2n+1)!} = frac{2n+1-1}{2(2n+1)!} = frac{1}{2 (2n)!} - frac{1}{2(2n+1)!}$

How to compute $sum_{n=1}^infty{frac{n}{(2n+1)!}}$?

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